about to only mathematics

Options

A. `x/5=(y-1)/8=(z-2)/(-13)`
B. `x/2(y-1)/3=(z-2)/(-5)`
C. `x/4=(y-1)/3=(z-2)/(-7)`
D. `x/2=(y-1)/(-7)=(z-2)/5`

Answer

D

Explanation

The coordinates of the any point on the line `(x+2)/2=(y+1)/(-1)=z/3` are given by `(x+2)/2=(y+1)/(-1)=z/3=r`
So the coordinats of any point on the given line are `P(2r-2,-4-1,3r)`.
Let `M(u,v,w)` be the feet of perpendicular drawn from `P` on the plane `x+y+z=3`. Then the direction ratios of `PM` are to the plane the direction ratios of whose normal are proportional to 1,1,1.
`:.(2r-2-u)/1=(-r-1-v)/1=(3r-w)/1=lamda`(say)
`impliesu-2r-2-lamda,v=-r-1-lamda,w=3r-lamda`
Since `M(u,v,w)` lies on the plane `x+y+z=3`. Therefore,
`u+v+w=3`
`implies2r-2-lamda-r-1-lamda+3-lamda=3`
`implies4r-3-3lamda=3implieslamda=(4r-6)/3`
`:.u=2r-2-(4r-6)/3,v=-r-1-(4r-6)/3,w=3r-(4r-6)/3`
`impliesu=(2r)/3,v=(-7r+3)3,w=(5r+6)/3`
`impliesu/(2//3),(v-1)/(-7//3)=(w-2)/(5//3)=r`
Hence `(u,v,w)` lies n the line
`x/(2//3)=(y-1)/(-7//3)=(w-2)/(5//3)` or `x/2=(y-1)/(-7)=(z-2)/5`